Q78. - Algebra 1

Question

Michael is mixing two types of candy for a party. The chocolate pieces cost$\mathrm{\backslash (}0.40$ per ounce, and the hard candy costs $\mathrm{\backslash )}0.20$ per ounce. Michael purchases $20$ ounces of the chocolate pieces, and the total cost of his candy was $\mathrm{\$}11$. How many ounces of hard candy did he purchase?

Step-by-Step Solution

Answer

The number of the ounces of the hard candy purchased is $15$.

Step 1. Given information.

The chocolate pieces cost $\mathrm{\$}0.40$ per ounce.

The hard candy costs $\mathrm{\$}0.20$ per ounce.

Michael purchases $20$ ounces of the chocolate pieces.

The total cost of his candy was $\mathrm{\$}11$.

Step 2. To find the expression of the problem.

Suppose the number of the ounces of the hard candy is $=a$.

Then according to the given statement is:

$20\times 0.4+a\times 0.2=11$

Step 3. To find the value of a .

To solve the above equation

$\begin{array}{l}20\times 0.4+a\times 0.2=11\\ 8+0.2a=11\\ 0.2a=11-8\\ 0.2a=3\\ a=\frac{3}{0.2}\\ a=15\end{array}$

Hence, the number of the ounces of the hard candy purchased is $15$.